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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Jul 2007 12:25:26 IST
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If a b and c are in AP prove that a2(b+c) b2(c+a) c2(a+b)
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3 Jul 2007 12:31:11 IST
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hey prove wat??
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3 Jul 2007 12:33:47 IST
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lol
i think u forgot to write the whole question!!!
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3 Jul 2007 13:29:53 IST
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Hi I have corrected the question. Sorry for the inconvenience If a b and c are in AP prove that a2(b+c) b2(c+a) c2(a+b) are in AP
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3 Jul 2007 13:31:14 IST
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are i understud the ques..u wanna me to prove tht the following three terms are in AP..simple see how..... b^2(c+a) - a^2(c+b)=c^2(a+b) - b^2(a+c)....if they r in ap => c(b^2 - a^2) + ab(b-a)=a(c^2 - b^2) + bc(c-b) => (b-a)(ab+bc+ca)=(c-b)(ab+bc+ca) =>b-a=c-b which is true as a b c are in ap
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a 2nd year IIT DELHI student, doing B.Tech in chemical engineering |
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3 Jul 2007 13:48:06 IST
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Thanks for your replies. I have a question. Can we prove that a2(b+c) b2(c+a) c2(a+b) are in AP if we start from a, b, and c which are also in AP
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3 Jul 2007 18:10:06 IST
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yes. as below. 2b = a + c (a,b, c are in AP) : now suaring : 4b^2 = a^2 + c^2 + 2.a.c. : Now dividing by 2 2b^2 = (a^2+c^2) /2 + (ac) : multiply both sides by (a+C) 2b^2 * (a+c) = (a^2 + c^2)*(a+c) / 2 + (ac)*(a+c) : rearranging : 2b^2 * (a+c) = (a^2 + c^2)*(2b)/2 + (ac)*(a+c) : exapnding now 2b^2 * (a+c) = b.a^2 + b.c^2 + c.a^2 + a.c^2 : rewriting again 2b^2 * (a+c) = b.a^2 + c. a^2 + b.c^2 + a. c^2 : Rearranging again 2b^2 * (a+c) = a^2 *(b+c) + c^2 ( a+b) : Hence, QED
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3 Jul 2007 18:30:03 IST
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hello frnd
obj approach
putting
a=1 b=2 c=3
so clearly a2(b+c) b2(c+a) c2(a+b) r also in a p
(5,16,27)
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3 Jul 2007 21:58:23 IST
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a,b,c r in AP
therefore, a^2 , b^2 , c^2 r also in AP
multiplying all 3 by a+b+c.....we get,
a^3 + a^2(b+c)
b^3 + b^2(a+c)
c^3+ c^2(a+b) r in AP
as a^3,b^3,c^3 r also in AP...WE CAN SUBSTRACT DAT FRM D EQUATION.....
AND V GET....
a^2(b+c),b^2(a+c) and c^2(a+b) r in AP.
I HOPE IT HELPS....
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we are the allies of konoha.....shinobis of the....sand
shubham sunder |
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4 Jul 2007 10:24:36 IST
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Am sorry to be a bit of spoilsport.
But if a,b,c are in AP, it does not necessarily imply that a^2, b^2, c^2 are in AP.
For eg, 1,3, 5 are in AP, but their swuares 1, 9, 25 are not in AP.
Rgds.
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5 Jul 2007 12:28:26 IST
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oldigol theres nothing to be spoilspot..u r rite..shubh has got it wrong....i have given the soln above..chek it out
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a 2nd year IIT DELHI student, doing B.Tech in chemical engineering |
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